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Ok, say I have a 2.5mm2 or 3.31mm2 copper conductor 90*C rated. 100,000 amps of available fault current. 277/480 volt Y system, TN-C-S. MCB protection. Would the conductor survive such a short circuit? How do I determine this?
 
As @Lister1987 says - do the adiabatic computation and you will find out.

But plugging in your values (2.5mm, 100kA and 90C copper so say k=100) leads to a disconnect time of 6.25 microseconds. So no, a MCB will not protect that cable.

But a MCB won't allow you to break 100kA either! You would be looking at the bigger end range of MCCB for that, or a fuse.

That is I2t of 62.5k A2s which is below the limiting case for BS88 fuses to around 80A limit so the cable can be protected (subject to impedance, as at low fault currents a fuse's I2t increases), but not be something with higher let-through. e.g. the Hager MCCB at 40kA PFC has a let-through of around 40M A2s, 640 times higher!

As an aside, you will find that fuses do compute to microsecond disconnection times but that is "virtual time" based on let-through and PFC^2, and not actual fusing time. Fuses generally will peak current limit, so the burn time is longer than the virtual time but what you normally want to know is the equivalent current/time graph so you can do tricks like determining selectivity or I2t limits graphically.
 
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But there is my conundrum, just a few feet of wire will limit the fault current significantly.
But that is your first step?

  1. Determine Zs at the end point (supply Ze and the cable resistances R1+R2) to get the PFC
  2. From the PFC and OCPD determine disconnect time
  3. From the conductor CAS, conductor material, and insulation temperature limits compute the I2t limit from the adiabatic equation
  4. Compare the resulting PFC^2 * time with the I2t limit for your conductor
 
But there is my conundrum, just a few feet of wire will limit the fault current significantly.

Not really where are you planning to have this unplanned fault?

At the far end with reduced fault level, or somewhere along its length?

If you assume the fault level is very close to the source - then determine the current, hence disconnect time, and then at the remote end and repeat the calculation you will cover whichever is the worst case.

Unless you accept that since the cable will need to be replaced in the event of a short fault, there is no use ensuring it will survive the heating effect anyhow, in which case only the far case would have to be considered and you just have to document why protection has not been applied.


I would also check what you mean by 100kA - is this prospective symmetrical RMS current or peak? - the adiabatic equation is based on RMS, whilst many US specifications use some form of peak.
 
But that is your first step?

  1. Determine Zs at the end point (supply Ze and the cable resistances R1+R2) to get the PFC
  2. From the PFC and OCPD determine disconnect time
  3. From the conductor CAS, conductor material, and insulation temperature limits compute the I2t limit from the adiabatic equation
  4. Compare the resulting PFC^2 * time with the I2t limit for your conductor


Right- however a few feet of wire bring down the current significantly. At 10 feet 3.31 mm2 will drop down to 13,850 amps assuming an infinite source.

If we assumed zero feet of wire in every instance we couldn't run anything smaller than 6mm2.
 
With 100kA PFC you have to seriously consider your fault protection and if it can meet the adiabatic limits of those cables. In all likelihood you are looking at more than one OCPD in cascade as somthing like 3-4mm is probably going to be a ciruit under 100A and not streight off the incoming OCPD, and need to be sure they can safely clear the fault as well as keeping the cable safe.

As @Julie. said above, you should check the computation for both ends (assuming faults likely at any point) as different PFC and the resulting different disconnection times and resulting I2t might show oe end is OK and not the other.
 
Right, however how much should I assume at the start of the run? Few feet? More, less? The thing is the NEC doesn't give any guidance, and BS7671 seems to ask using fault current levels that are not likely to be encountered in reality ie I can't see a wire shorting to the enclosure 2 inches off the MCB.
 
Right, however how much should I assume at the start of the run? Few feet? More, less? The thing is the NEC doesn't give any guidance, and BS7671 seems to ask using fault current levels that are not likely to be encountered in reality ie I can't see a wire shorting to the enclosure 2 inches off the MCB.
Thing is you can get a short inches from the MCB/MCCB if someone failed to fit a grommet and the cable is cut by a metal enclosure's edge!

Having 100kA available is very high (at least in my books). I would expect that to be split and fed to various sub-boards where the smaller loads are, and by that point you might be down at a PFC that is easier to deal with and let-through that will be OK for the adiabatic limits of any cables.

Why so thin a conductor if the OCPD can't protect it?

It seems the logical solutions are either (1) a thicker wire that can survive the fault-clearing, and/or (2) make sure the choice of OCPD is able to both break 100kA PFC and to have a let-through low enough not to cause a fire, etc.

As an aside, my current project is fed off a local 500kVA transformer and at the point we get it (about 20m from the transformer) there is no more than 12kA bolted fault current (perhaps a few kA more if any of the large motors are running at the time and act as generators dueing the fault, of course). As we are using 100A BS88 fuses in a fused-switch of the busbar chamber we pretty much don't have to care as the let-through is below what any of the downstream MCB can safely interrupt.
 

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